# Show that if $R$ is a division ring and $S$ is a subring of $R$ then $S$ is also a division ring.

I'm trying to do problem 7.1.7 in Dummit and Foote.

Problem: Prove that the centre of a ring $$R$$ is a subring of $$R$$ that contains the identity. Prove that the center of a division ring is a field.

I've already shown that $$Z(R)$$ is a subring, but my issue here is that I would like to show that if $$S \subseteq R$$ is a subring of a division field then $$S$$ must itself be a division ring. This would then show that the centre of a division ring is a field since it will be a commutative division ring. However I don't seem to be able to finish it off even though it seems basic. I know that an additive subgroup of $$R$$ that is also closed under multiplication and contains the multiplicative identity, but I can't seem to leverage this to show that any element in $$S$$ has a multiplicative inverse. As I'm sure it's basic a hint would be appreciated, thanks in advance for your help.

• It is not true that any subring of a division ring is itself a division ring. Consider the integers as a subring of the rationals. Dec 5, 2022 at 19:37
• Oh of course sorry! Thanks @MichaelCohen. If you'd leave that as an answer I would accept it to clear it from unanswered. Up to you. Dec 5, 2022 at 19:39

The key to the problem you ask in the body (not the title) is that if $$x \in Z(R)$$ and $$x^{-1} \in R$$, then $$x^{-1} \in Z(R)$$ also. Notice for any $$r \in R$$ and $$x \in Z(R)$$, we have $$xr = rx$$ so $$x^{-1}(xr)x^{-1} = x^{-1}(rx)x^{-1},$$ and, by re-associating and simplifying, we get $$rx^{-1}=x^{-1}r$$ for all $$r \in R$$. Thus $$x^{-1} \in Z(R)$$.
Thus $$Z(R)$$ is a commutative subring of $$R$$ closed under taking inverses and is thus a commutative division ring, i.e. a field.